> Erm, no they don't. If your "for every n there exists k such that" was instead "there exists k such that for every n" then they would contradict each other.

Try to find a d_n that, together with all its predecessors d_j for 0 < j < n,is not in a q_k. Then you may boast "erm". Otherwise learn that, in mathematics, there is no continuation of d beyond every d_n.